The capacity criterion above requires the obstacle envelope of $u_0$ in a smaller background $\delta \omega $ not to be identically $-\infty $. We give two sufficient conditions of different origin: invariance of the Monge–Ampère mass controls the contact measures of envelopes of stretched truncations, while a sublinear bound on the singularity provides explicit competitors. Both conditions also give zero Lelong numbers. In the setting of theorem 12.10.17, if the maximizing form has a globally smooth root of rank at least two, zero Lelong numbers then allow the threshold solution to be extended smoothly across the analytic degeneracy set.
The setting is unchanged: $X$ has dimension $n\geq 2$, $\omega $ is a Hermitian form and $\mathrm{d}V$ a smooth positive probability volume form. Inequalities between quasi-plurisubharmonic functions refer to their canonical representatives, and $\nu (u,x)$ denotes the Lelong number of $u$ at $x$. We will use the following consequence of lemma 5.38.8, where $C_\omega $ is the bound for the Lelong numbers of $\omega $-plurisubharmonic functions given there: if $u\in \mathrm{PSH}(X,\omega )$ and, for every $0<\delta <1$, some $v_\delta \in \mathrm{PSH}(X,\delta \omega )$ satisfies $v_\delta \leq u$, then
\[ 0\leq \nu (u,x)\leq \nu (v_\delta ,x) =\delta \nu (v_\delta /\delta ,x)\leq \delta C_\omega . \]
Thus $u$ has zero Lelong numbers; the bound is uniform in $\delta $ because $v_\delta /\delta \in \mathrm{PSH}(X,\omega )$.
Theorem 12.10.23.30380C32B7A7 Suppose $\mathrm{dd}^{\mathrm{c}}(\omega ^ j)=0$ for $1\leq j<n$. Put $V=\int _ X\omega ^ n$. For every nonzero $0\leq f\in L^1(\mathrm{d}V)$, with $M=\int _ Xf\, \mathrm{d}V$, one has
\[ b_\omega (f)=\frac VM. \]
Every threshold subsolution $u$ satisfies $\mathrm{MA}^{\mathrm{ext}}_\omega (u)=(V/M)f\, \mathrm{d}V$ and has mass $V$. Its obstacle envelope $P_{\delta \omega }(u)$ is not identically $-\infty $ for any $0<\delta <1$, and $u$ has zero Lelong numbers. More generally, these last two conclusions hold for every $u\in \mathrm{PSH}(X,\omega )$ with $\int _ X\mathrm{MA}^{\mathrm{ext}}_\omega (u)=V$.
Proof.
By theorem 11.1.7, every bounded $\omega $-plurisubharmonic function has mass $V$, and truncation therefore bounds every extended mass by $V$. For $f_ k=\min (f,k)$, integration of the bounded equation gives $c_\omega (f_ k)=V/\int _ Xf_ k\, \mathrm{d}V$, so the definition (15.28.1) gives $b_\omega (f)=V/M$. A threshold subsolution has mass at least $b_\omega (f)M=V$. Its mass is consequently $V$, and the nonnegative difference between its measure and $b_\omega (f)f\, \mathrm{d}V$ has mass zero.
Now let $u\in \mathrm{PSH}(X,\omega )$ have extended mass $V$. Normalize $u\leq 0$, put $\mu _ u=\mathrm{MA}^{\mathrm{ext}}_\omega (u)$, and fix $a>1$. For $u_ j=\max (u,-j)$, let $w_ j=P_\omega (au_ j)$. Comparison with constants gives $-aj\leq w_ j\leq 0$. The bounded contact theorem theorem 16.28.4 shows that $\mathrm{MA}_\omega (w_ j)$ is concentrated on the contact set $C_ j=\{ w_ j=au_ j\} $. Since the two functions are canonical representatives and $w_ j\leq au_ j$ almost everywhere, this inequality holds everywhere. Both functions are $a\omega $-plurisubharmonic, so positivity and proposition 13.14.6 give
\[ \mathrm{MA}_\omega (w_ j) \leq \mathbf1_{C_ j}\mathrm{MA}_{a\omega }(w_ j) \leq a^ n\mathbf1_{C_ j}\mathrm{MA}_\omega (u_ j). \]
For integers $j\geq N$, locality of the products on $\{ u>-N\} $ and the mass identity for bounded functions yield
\[ \int _{\{ u_ j\leq -N\} }\mathrm{MA}_\omega (u_ j) =V-\mu _ u(\{ u>-N\} )=\mu _ u(\{ u\leq -N\} ). \]
Choose $N$ with $a^ n\mu _ u(\{ u\leq -N\} )<V$; this is possible because $\mu _ u$ is finite and puts no mass on the pole set of $u$. If $\sup _ Xw_ j<-aN$, the contact set $C_ j$ lies in $\{ u_ j<-N\} $, and the preceding inequalities contradict $\int _ X\mathrm{MA}_\omega (w_ j)=V$. Hence $\sup _ Xw_ j\geq -aN$ for $j\geq N$, and for smaller $j$ the bound $\sup _ Xw_ j\geq -aj$ suffices.
The envelopes $w_ j$ decrease with $j$, and by compactness of normalized quasi-plurisubharmonic functions this uniform lower bound for their suprema prevents their limit from being identically $-\infty $. Their limit $w\in \mathrm{PSH}(X,\omega )$ satisfies $w\leq au$, also at the poles of $u$. With $\delta =1/a$, the function $\delta w\in \mathrm{PSH}(X,\delta \omega )$ is therefore a competitor for $P_{\delta \omega }(u)$, so this envelope is not identically $-\infty $. Since $a>1$ was arbitrary, this holds for every $0<\delta <1$, and the Lelong estimate preceding the theorem completes the proof. The normalization $u\leq 0$ does not affect any conclusion.
$\square$
Proposition 12.10.24.62BDDC322780 Let $\psi \leq 0$ be a quasi-plurisubharmonic function with $\mathrm{dd}^{\mathrm{c}}\psi \geq -C\omega $ for some $C>0$, and let $u\in \mathrm{PSH}(X,\omega )$ satisfy, everywhere on $X$,
\[ u\geq -A(1-\psi )^\alpha -B, \qquad A>0,\quad B\in \mathbb R,\quad 0<\alpha <1. \]
Then $P_{\delta \omega }(u)$ is not identically $-\infty $ for every $0<\delta <1$, and $u$ has zero Lelong numbers.
Proof.
Choose $\alpha <\beta <1$ and $\varepsilon >0$ with $\varepsilon \beta C\leq \delta $. The number
\[ K=B+\sup _{t\geq 1}(At^\alpha -\varepsilon t^\beta ) \]
is finite, and $v_\delta =-\varepsilon (1-\psi )^\beta -K\leq u$. The function $t\mapsto -\varepsilon (1-t)^\beta $ is increasing and convex on $t\leq 0$, with derivative at most $\varepsilon \beta $. Applying the chain inequality for quasi-plurisubharmonic functions to $\max (\psi ,-j)$ and passing to the decreasing limit as $j\to \infty $, we obtain $\mathrm{dd}^{\mathrm{c}}v_\delta \geq -\varepsilon \beta C\omega \geq -\delta \omega $. The limit is not identically $-\infty $ because $\psi $ is finite almost everywhere; at the poles of $\psi $ it takes the value $-\infty $, where the comparison with $u$ still holds. Thus $v_\delta \in \mathrm{PSH}(X,\delta \omega )$ is a competitor for $P_{\delta \omega }(u)$ for each $\delta $, and the Lelong estimate preceding theorem 12.10.23 proves the last assertion.
$\square$
Pome
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